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25 January, 02:30

Consider the initial value problem 25y′′+40y′+16y=0, y (0) = a, y′ (0) = -1. Find the critical value of a that separates solutions that become negative from those that are always positive for t>0.

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  1. 25 January, 02:53
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    Step-by-step explanation:

    Given the differential equation

    25y′′+40y′+16y=0

    Using D operator to find the complementary solution

    Since the Differential equation is equal to 0, then it doesn't have a partial solution.

    25y′′+40y′+16y=0

    25D² + 40D + 16 = 0

    25D² + 20D + 20D + 16 = 0

    5D (5D+4) + 4 (5D+4) = 0

    (5D+4) = 0 twice

    D = - 4/5 twice,

    So the solution is a real and equal roots

    y (t) = (A+B•t) exp (-0.8t)

    Where A and B are constant

    The initial value are.

    y (0) = a, y′ (0) = -1

    y (0) = a = (A+B (0)) exp (-0.8 (0))

    a = A

    Then, the constant A = a.

    Now, y' (t)

    y' (t) = B•exp (-0.8t) - 0.8 (A+B•t) exp (-0.8t

    -1 = B - 0.8A

    Since A = a

    Then, B = 0.8a - 1

    The solution becomes

    y (t) = (a + (0.8a-1) •t) exp (-0.8t)

    For t>0

    For positive,

    a + (0.8a - 1) > 0

    1.8a > 1

    a > 1/1.8

    a > 10/18 > 5/9

    a > 5/9.
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